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#120 by Monomorphic on 13 Mar, 2016 12:41
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Optical polishing almost done. Before and after pics of small endplate of NSF-1701A
Dave, is the thin concentric dark ring at the copper perimeter a groove for the side-walls to sit in?
Looks like your new copper is much thicker than 1mm.
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#121 by rfmwguy on 13 Mar, 2016 13:30
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It is a rough cut solder channel a mil or so deep. The cone will be formed and seam soldered then put into the channel. The disk will sit on a hotplate and full of reflowed solder, basically, a liquid moat of silver solder. The hotplate gets turned off and hopefully a uniform solder seal occurs. Slow temp changes should maintain flatness.
I've not done this before, thought I'd try something new for the fun of it.
Also, the disk is 1/8 inch thick. The large end will be the same. Figured warping would be minimal and I don't have to mess with gluing thin copper on a batting material. Expansion coefficients of different materials can be a problem at times. -
#122 by rfmwguy on 13 Mar, 2016 13:37
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Very nice. I meant to video the process, but got lazy. Copper is a weird metal, its softness really trashes tools and fine grit sandpaper. Found the best polish is 9, 3 and 1 micron diamond polish available in a syringe.Optical polishing almost done. Before and after pics of small endplate of NSF-1701A
Nice polishing job. You can use it as a faux bronze age mirror. I did something similar a few weeks ago. The brass disk (before and after) is the cover plate for a diesel engine water pump. Instead of buying a replacement I machined it flat then polished it. Later I found 2 new cover plates in a locker. -
#123 by Monomorphic on 13 Mar, 2016 13:53
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It is a rough cut solder channel a mil or so deep.
I hope you accounted for that 1mm in your side walls!
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#124 by rfmwguy on 13 Mar, 2016 14:05
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Yes, the cone is not cut yet. I'll oversize it then use the lapping plate to sand the cone down to a precise measurement between top and bottom plates. I have some ideas to frequency tune magnetron to match to whatever final resonance will be. Size of frustum will be fixed but think I have the mag fine tuning solution worked out.
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#125 by Monomorphic on 13 Mar, 2016 15:23
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Yes, the cone is not cut yet.
What thickness copper side walls? The 1/8 or sticking with 1mm? -
#126 by rfmwguy on 13 Mar, 2016 16:24
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1mm sidewalls. Was going to to spin thicker copper but no spinners responded with decent prices. A fully funded institutional project could spin then polish. A user here also suggested a lost wax pocess.Yes, the cone is not cut yet.
What thickness copper side walls? The 1/8 or sticking with 1mm? -
#127 by zen-in on 13 Mar, 2016 16:56
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1mm sidewalls. Was going to to spin thicker copper but no spinners responded with decent prices. A fully funded institutional project could spin then polish. A user here also suggested a lost wax pocess.Yes, the cone is not cut yet.
What thickness copper side walls? The 1/8 or sticking with 1mm?
Spinning is your best option. Copper is not amenible to the lost wax casting process. It does not liquify well in the presence of Oxygen. The outer surface converts to cupric oxide. Copper is cast in a vacuum, using graphite molds. -
#128 by Rodal on 13 Mar, 2016 18:10
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1mm sidewalls. Was going to to spin thicker copper but no spinners responded with decent prices. A fully funded institutional project could spin then polish. A user here also suggested a lost wax pocess.Yes, the cone is not cut yet.
What thickness copper side walls? The 1/8 or sticking with 1mm?
It is not going to be possible to have a high Q (quality of resonance) close to theoretical with 1 mm walls: very compliant (the opposite of stiff). For a length of 0.26 meters and diameter of 0.28 meters, a 1 mm wall thickness is easy to deform out of shape just by applying hand pressure, and hence difficult to maintain geometrical tolerance. -
#129 by SeeShells on 13 Mar, 2016 18:36
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1mm sidewalls. Was going to to spin thicker copper but no spinners responded with decent prices. A fully funded institutional project could spin then polish. A user here also suggested a lost wax pocess.Yes, the cone is not cut yet.
What thickness copper side walls? The 1/8 or sticking with 1mm?
It is not going to be possible to have a high Q (quality of resonance) close to theoretical with 1 mm walls: very compliant (the opposite of stiff). For a length of 0.26 meters and diameter of 0.28 meters, a 1 mm wall thickness is easy to deform out of shape just by applying hand pressure, and hence difficult to maintain geometrical tolerance.
I hope you don't mind me adding my .02 cents on a 1mm or .039" copper wall. Once the endplates are secured by soldering you could literally stand on the frustum, I did. That's not the biggest issue we face in deforming, it's the thermal aspect of deformation that is the Q killer.
Shell -
#130 by meberbs on 13 Mar, 2016 19:06
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Here is a simple excel calculator to show how many reflections will occur from each end plate, depending on unloaded cavity Q 5 TC time, Rf freq and number of 1/2 resonant waves in the cavity.
Example attached shows that for a TXXX3 resonant mode at 2.45 GHZ with a unloaded cavity Q of 86.2k, there would be 45,731 reflection from each end plate, for every cycle of Rf input, until the Rf energy is totally thermalised or converted into an externally usable force over 27.998usec. Which is a good life time for 1 cycle of 2.45GHz input Rf energy.
45,731 reflections from each end plate is a lot to keep perfectly aligned and for them to not walk off to a wider separation point. Of course the end plate alignment also affects cavity Q as it affects wall losses.
At least now we have a way to calc how many reflection will occur from each end plate if we get the cavity built perfect to 10x skin depth accuracy or better.
Urm, I don't suppose anybody can hack a simulator to check this to see if bounces off the sidewalls cause one endplate to undergo more bounces than the other. While CoM says it shouldn't matter, it would be nice to have the data to make sure nothing odd is going on.
"Number of bounces" has limited meaning in this context. It only makes sense when thinking about individual photons, and that does not provide useful insight to a resonator like this. For a 2 mirrors situation, you can calculate number of photons per time hitting each mirror easily, but the interactions with the side walls make such calculations less direct and meaningful.
The existence of individual photons is a quantum mechanical effect, and is not noticeable at macro scale/energies except when there is something such as the photoelectric effect where the energy delivered to a single electron needs to be a certain value to cause the effect. To give an idea, 1 Joule of energy is equivalent to over 600000000000000000000000 photons at 2.5 GHz (that is 23 zeros, I would use scientific notation but I want to emphasize this is a ridiculously huge number). You would need over 68 billion Terabytes of memory to have just one bit dedicated to each of those photons. Additionally since photons are overlapping quantum mechanical particles, it is literally meaningless and impossible to follow the path of any one photon.
Inside a frustum resonator, the fields are effectively the result of photons traveling in multiple directions where the fields are partially cancelling, generating the interesting patterns for different modes. I don't believe there is any way to easily reconstruct what different directions are underlying the patterns, if that even is a meaningful concept. Energy is being exchanged between the fields and the currents and charge distributions in the walls. Energy is lost as heat when the currents encounter resistance. This happens throughout the walls as a function of current density and is not directly tied to any discrete photon. Q under the definition being used here describes how quickly this energy turns into heat averaged over a cycle. -
#131 by TheTraveller on 13 Mar, 2016 21:01
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"Number of bounces" has limited meaning in this context....
Put an E field sensor through one end plate plus a lot of attenuation so it doesn't add significant load to the cavity.
Set up to measure the unloaded cavity Q and the number of positive peak events in the E field during 5 TCs of the cavity discharge time.
Fill the cavity with resonant Rf.
Stop the Rf input just as the Rf crosses zero.
Measure the time until the E field probe says there is no more Rf energy inside the cavity. Should be 5x TC.
Count the number of E field positive peak events during the 5 x TC time period (should occur at the rate of the Rf resonant freq). Should be the indicated end plate reflection count number. -
#132 by TheTraveller on 13 Mar, 2016 21:23
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Yes, the cone is not cut yet
If you still plan on exciting using the maggie antenna inserted in the middle of one end plate, expect to excite TM113 and not TE013 (same guide wavelength), which would require the antenna inserted through the side wall at one of the 3 peak E field locations.
Maybe one of the FEKO guys can model your build and antenna placement so you get the best cone end plate to end plate length to match resonance at your excitation freq and you understand what the TM113 mode fields should look like.
Best of luck. -
#133 by rq3 on 13 Mar, 2016 21:37
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"Number of bounces" has limited meaning in this context....
Put an E field sensor through one end plate plus a lot of attenuation so it doesn't add significant load to the cavity.
Set up to measure the unloaded cavity Q and the number of positive peak events in the E field during 5 TCs of the cavity discharge time.
Fill the cavity with resonant Rf.
Stop the Rf input just as the Rf crosses zero.
Measure the time until the E field probe says there is no more Rf energy inside the cavity. Should be 5x TC.
Count the number of E field positive peak events during the 5 x TC time period (should occur at the rate of the Rf resonant freq). Should be the indicated end plate reflection count number.
What the heck is this constant of 5 you keep using? It first reared it's head when you claimed that a wall thickness of 5 skin depths would yield zero field. Which isn't true. The strength is down to (off the top of my head) about 0.7%, (curiously close to 0.707?) but NOT zero.
It's similar to radioisotope half-lives. The radiation just gets statistically very small, until it's effectively negligable. Making arbitrary statements about what is or is not negligable while studying an unknown and unproved effect (the Emdrive) is...odd. IMHO. -
#134 by rq3 on 13 Mar, 2016 21:44
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1mm sidewalls. Was going to to spin thicker copper but no spinners responded with decent prices. A fully funded institutional project could spin then polish. A user here also suggested a lost wax pocess.Yes, the cone is not cut yet.
What thickness copper side walls? The 1/8 or sticking with 1mm?
Dave, if you're referring to electroforming as a lost wax process, it is not. The form is often reusable, and needn't be made of wax. It just seems to me that machinable wax would be an easy and inexpensive material for the DIY community, with access to a lathe, to fabricate an accurate frustum with excellent surface finish.
If you're referring to some comment I missed regarding actual lost wax casting, my apologies. I agree that lost wax casting (as in pouring molten copper) of a frustum would be an exercise in futility for the DIY group. Brass or bronze could work, though, if you were willing to do quite a bit of post casting finishing. -
#135 by TheTraveller on 13 Mar, 2016 21:50
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"Number of bounces" has limited meaning in this context....
Put an E field sensor through one end plate plus a lot of attenuation so it doesn't add significant load to the cavity.
Set up to measure the unloaded cavity Q and the number of positive peak events in the E field during 5 TCs of the cavity discharge time.
Fill the cavity with resonant Rf.
Stop the Rf input just as the Rf crosses zero.
Measure the time until the E field probe says there is no more Rf energy inside the cavity. Should be 5x TC.
Count the number of E field positive peak events during the 5 x TC time period (should occur at the rate of the Rf resonant freq). Should be the indicated end plate reflection count number.
What the heck is this constant of 5 you keep using? It first reared it's head when you claimed that a wall thickness of 5 skin depths would yield zero field. Which isn't true. The strength is down to (off the top of my head) about 0.7%, (curiously close to 0.707?) but NOT zero.
It's similar to radioisotope half-lives. The radiation just gets statistically very small, until it's effectively negligable. Making arbitrary statements about what is or is not negligable while studying an unknown and unproved effect (the Emdrive) is...odd. IMHO.
At 1x skin depth the surface current density is down to approx 37% or 1/e. Takes 5x skin depth to reduce it to approx 100%.
For a resonant cavity 1x TC is unloaded Q / (2 × Pi × freq resonant). Takes 5x that time to fully fill or discharge the cavity. 1x TC time will discharge the cavity to 1/e or ~37%.
Please see my earlier post:
http://forum.nasaspaceflight.com/index.php?topic=39772.msg1502999#msg1502999
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#136 by Rodal on 13 Mar, 2016 22:03
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1mm sidewalls. Was going to to spin thicker copper but no spinners responded with decent prices. A fully funded institutional project could spin then polish. A user here also suggested a lost wax pocess.Yes, the cone is not cut yet.
What thickness copper side walls? The 1/8 or sticking with 1mm?
It is not going to be possible to have a high Q (quality of resonance) close to theoretical with 1 mm walls: very compliant (the opposite of stiff). For a length of 0.26 meters and diameter of 0.28 meters, a 1 mm wall thickness is easy to deform out of shape just by applying hand pressure, and hence difficult to maintain geometrical tolerance.
I hope you don't mind me adding my .02 cents on a 1mm or .039" copper wall. Once the endplates are secured by soldering you could literally stand on the frustum, I did. That's not the biggest issue we face in deforming, it's the thermal aspect of deformation that is the Q killer.
Shell
Perhaps what I wrote was not readily understandable, so I will try to explain it further. The issue I was concerned with is the issue of tolerance that DIY people should use. TheTraveller quoted Shawyer as invoking a tolerance of 13 micrometers (which to me is way too stringent for 2.45 GHz microwaves with a free-space wavelength of 0.122 m). On the other hand, it appears that rfmwguy and SeeShells have a much, much loser tolerance.
With a shell structure that is 1 mm thick but 0.28 m diameter and 0.26 m long the issue is deformation, not strength.
Stiffness is not the same thing as strength.
Strength is the highest stress that a structure can carry before it fails (it can be defined as permanent deformation (*) for a tough material or fracture for a brittle material).
On the other hand, stiffness is the ratio of strain (change in length per unit original length) to stress, or the ratio of deformation by force.
A structure can be strong but not stiff enough for a given application.
The issue with the quality of resonance is not at all anything to do with strength.
When you say:QuoteOnce the endplates are secured by soldering you could literally stand on the frustum, I did.
you may be able to determine just by standing on it is whether you exceeded the strength of the structure, certainly if it fractures, or perhaps if it permanently deforms and the permanent deformation is large enough to be perceived.
To check the stiffness, you would need to be able to apply a load (stand on it) and simultaneously measure the deformation while the load is being applied. This is usually done with universal testers (either mechanical, screw-driven, or servo-hydraulic testers).
For elastic deformation, the structure will return to its original shape once you unload from it.
The concern is that the Q quality of deformation will be affected by deformations of the order of a mm.
There is a huge dissonance between TheTraveller writing that Shawyer is asking for a tolerance of 13 micrometers in order to achieve a Q close to theoretical and on the other hand, saying that all the tolerance needed is whether one can observe deformation of truncated cone just by standing on it.
I don't think that you would be able to determine an elastic deformation of a mm just by standing on it (you would need to have mirrors and an incredible vision to be able to tell a mm deformation !!! ). (*)
So what is needed here is to quantify stiffness: to quantify deformation of the structure. Even when looking for permanent deformation, since the expected permanent deformation is of the order of a mm (the wall thickness) (**) standing on the truncated cone may not be an accurate way to determine it..
And what is needed is for DIY experimenters to agree on a tolerance: there is a huge gap between the 13 micrometer tolerance invoked by TheTraveller/Shawyer and on the other hand invoking that if one can stand on the truncated cone, that is good enough.
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(*) Actually, one does not need to really stand on the structure to be able to figure out the deformation of a cylinder that is 0.26 m long by 0.28 m diameter with a wall thickness of only 1 mm . One can readily use the theory of elasticity for thin shells to figure the deformation of cylindrical shell under load
. The problem, though, is what is the boundary condition fixing the end plates to the cylindrical section: is it simply supported (appropriate for thin end plates) or is it cantilevered ends (which would be appropriate for very thick end plates).
(**) The "standing on the cone" structure did serve as a test that you did not reach snap-through permanent buckling, but permanent deformation of the order of thickness (mm) was still certainly possible.
The Q of the truncated cone will be affected by deformations much smaller than something like the snap-through buckling permanent deformation shown in this picture :
It would be good for DIY to agree on a tolerance that may not be as extreme as the 13 micrometers invoked by TheTraveller/Shawyer, but yet is not as loose as "standing on it" -
#137 by rfmwguy on 13 Mar, 2016 22:08
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Yes, electroforming using wax that can be melted and drained out of the rf insertion port. Sorry, just reminded me of lost wax which is molten metal.
1mm sidewalls. Was going to to spin thicker copper but no spinners responded with decent prices. A fully funded institutional project could spin then polish. A user here also suggested a lost wax pocess.Yes, the cone is not cut yet.
What thickness copper side walls? The 1/8 or sticking with 1mm?
Dave, if you're referring to electroforming as a lost wax process, it is not. The form is often reusable, and needn't be made of wax. It just seems to me that machinable wax would be an easy and inexpensive material for the DIY community, with access to a lathe, to fabricate an accurate frustum with excellent surface finish.
If you're referring to some comment I missed regarding actual lost wax casting, my apologies. I agree that lost wax casting (as in pouring molten copper) of a frustum would be an exercise in futility for the DIY group. Brass or bronze could work, though, if you were willing to do quite a bit of post casting finishing. -
#138 by SeeShells on 13 Mar, 2016 23:16
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1mm sidewalls. Was going to to spin thicker copper but no spinners responded with decent prices. A fully funded institutional project could spin then polish. A user here also suggested a lost wax pocess.Yes, the cone is not cut yet.
What thickness copper side walls? The 1/8 or sticking with 1mm?
It is not going to be possible to have a high Q (quality of resonance) close to theoretical with 1 mm walls: very compliant (the opposite of stiff). For a length of 0.26 meters and diameter of 0.28 meters, a 1 mm wall thickness is easy to deform out of shape just by applying hand pressure, and hence difficult to maintain geometrical tolerance.
I hope you don't mind me adding my .02 cents on a 1mm or .039" copper wall. Once the endplates are secured by soldering you could literally stand on the frustum, I did. That's not the biggest issue we face in deforming, it's the thermal aspect of deformation that is the Q killer.
Shell
Perhaps what I wrote was not readily understandable, so I will try to explain it further.
Stiffness is not the same thing as strength.
Strength is the highest stress that a structure can carry before it fails (it can be defined as permanent deformation for a tough material like steel or fracture for a brittle material like a ceramic).
On the other hand, stiffness is the ratio of strain (change in length per unit original length) to stress.
The issue with the quality of resonance is not at all anything to do with strength.
When you say:QuoteOnce the endplates are secured by soldering you could literally stand on the frustum, I did.
all you may be able to determine is whether you exceeded the strength of the structure, certainly if it fractures, or perhaps if it permanently deforms and the permanent deformation is large enough to be perceived.
To check the stiffness, you would need to be able to apply a load (stand on it) and simultaneously measure the deformation while the load is being applied. For elastic deformation, the structure will practically instantaneously return to its original shape once you unload from it.
In this case, one can readily show that the Q quality of deformation will be severely affected by deformations of the order a mm.
I don't think that you would be able to determine an elastic deformation of a mm just by standing on it (you would need to have mirrors and an incredible vision to be able to tell a mm deformation !!! ).
And one does not need to really stand on the structure to be able to figure out the deformation of a cylinder that is 0.26 m long by 0.28 m diameter with a wall thickness of only 1 mm . One can readily use the theory of elasticity for thin shells to figure the deformation of the truncated cone under load.
So what is needed here is to quantify stiffness: to quantify deformation of the structure. Even when looking for permanent deformation, since the expected permanent deformation is of the order of a mm (the wall thickness) (*) standing on the truncated cone may not be an accurate way to determine it.
_____
(*) The "standing on the cone" structure did serve as a test that you did not reach snap=through permanent buckling, but permanent deformation of the order of thickness (mm) was still certainly possible.
Are we talking about two different things Dr. Rodal? The Young's modulus verses thermal expansion coefficient of copper. If your talking about copper being changed from the pressures of a hand in the copper frustum you would have to exert.
Longitudinal stress = Force(F)/Cross sectional area (A) = F/A
Longitudinal strain = Extension(e)/Original length (Lo) = e/Lo
(E) = [F/A]/[e/Lo] = FLo/eA
Copper=120 Gpa (Young modulus)
Whereas putting it into perspective.
Iron=110 Gpa
For the frustum to show deviations from the pressures of just the hand and a significant change in the endplate Z distance of 1mm (where tuning is the most critical) you would have to exert hundreds of pounds.
What they are more likely seeing is a thermal expansion of the copper from the heat in the hand on the copper causing a expansion and the endplate distances to change. Or it's likely we are changing the harmonic frequency from the capacitance added from the hand on the walls.
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#139 by Rodal on 13 Mar, 2016 23:37
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Are we talking about two different things Dr. Rodal? The Young's modulus verses thermal expansion coefficient of copper. If your talking about copper being changed from the pressures of a hand in the copper frustum you would have to exert.
Longitudinal stress = Force(F)/Cross sectional area (A) = F/A
Longitudinal strain = Extension(e)/Original length (Lo) = e/Lo
(E) = [F/A]/[e/Lo] = FLo/eA
Copper=120 Gpa (Young modulus)
Whereas putting it into perspective.
Iron=110 Gpa
For the frustum to show deviations from the pressures of just the hand and a significant change in the endplate Z distance of 1mm (where tuning is the most critical) you would have to exert hundreds of pounds.
What they are more likely seeing is a thermal expansion of the copper from the heat in the hand on the copper causing a expansion and the endplate distances to change. Or it's likely we are changing the harmonic frequency from the capacitance added from the hand on the walls.
No, we are not dealing with a solid bar of copper here, instead we are dealing with thin shell structures.
One way to simplify the problem of what is the stiffness of a cylindrical shell is to look at the change in diameter of a cylindrical shell with open ends, under the action of a uniform line load, for which there is an easy formula.
Change in diameter=(1 - nu^2)*1.78853*((Load per unit length)/ E) *(Radius/thickness) ^3
(Observe that the stiffness of a cylindrical shell is proportional to the cube of the ratio of the thickness to the diameter, so that thickness is an extremely important variable, going like the third power !!!)
for:
Radius=(0.28 m)/2
Thickness = 0.001 m
E = 17*10^6 psi = 117 GPa
Change in diameter = 0.001 m
Poisson's ratio= nu = 0.3
requires a force/unitLength = 0.15 lbf/in (0.15 pounds force per inch)
or just 1 pound force distributed every 6.7 inches of length
The calculation for solid ends is more complicated, but you get the idea, I hope.
. The problem, though, is what is the boundary condition fixing the end plates to the cylindrical section: is it simply supported (appropriate for thin end plates) or is it cantilevered ends (which would be appropriate for very thick end plates).